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Chapter 9: Q28P (page 230)

(III) You are on a pirate ship and being forced to walk the plank (Fig. 9–68). You are standing at the point marked C. The plank is nailed onto the deck at point A, and rests on the support 0.75 m away from A. The center of mass of the uniform plank is located at point B. Your mass is 65 kg and the mass of the plank is 45 kg. What is the minimum downward force the nails must exert on the plank to hold it in place?

Short Answer

Expert verified

The minimum downward force that the nails must exert on the plank to hold it in place is \(2352\;{\rm{N}}\).

Step by step solution

01

Meaning of torque

The torque value, mathematically, can be determined by multiplying the value of the applied force and the distance of the force from a pivoted point.

02

Given information

Given data:

The mass of the person is \(M = 65\;{\rm{kg}}\).

The mass of the plank is \(m = 45\;{\rm{kg}}\).

03

Free body diagram of the plank

The free body diagram of the plank can be drawn as:

Here, \({F_{{\rm{nails}}}}\) is the force exerted on the nails and \({F_{{\rm{pivot}}}}\) is the force exerted on the pivot point.

04

Evaluation of the minimum downward force that the nails must exert on the plank to hold it in place

Take torques about the pivot point, with the clockwise torques, as positive.

\(\begin{array}{c}\sum \tau = 0\\\left\{ \begin{array}{l}mg\left( {0.75\;{\rm{m}}} \right)\cos \theta + Mg\left( {2.25\;{\rm{m}}} \right)\cos \theta \\ - {F_{{\rm{nails}}}}\left( {0.75\;{\rm{m}}} \right)\cos \theta \end{array} \right\} = 0\\{F_{{\rm{nails}}}} = \frac{{mg\left( {0.75\;{\rm{m}}} \right)\cos \theta + Mg\left( {2.25\;{\rm{m}}} \right)\cos \theta }}{{\left( {0.75\;{\rm{m}}} \right)\cos \theta }}\\{F_{{\rm{nails}}}} = \frac{{mg\left( {0.75\;{\rm{m}}} \right) + Mg\left( {2.25\;{\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}{F_{{\rm{nails}}}} = \frac{{\left( {45\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {0.75\;{\rm{m}}} \right) + \left( {65\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2.25\;{\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\{F_{{\rm{nails}}}} = 2352\;{\rm{N}}\end{array}\)

Thus, the minimum downward force that the nails must exert on the plank to hold it in place is \(2352\;{\rm{N}}\).

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Most popular questions from this chapter

(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door’s weight (Fig. 9–69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

(II) (a) Calculate the magnitude of the force, required of the “deltoid” muscle to hold up the outstretched arm shown in Fig. 9–72. The total mass of the arm is 3.3 kg. (b) Calculate the magnitude of the force exerted by the shoulder joint on the upper arm and the angle (to the horizontal) at which it acts.

When you apply the torque equation \(\sum {\tau = 0} \) to an object in equilibrium, the axis about which the torques are calculated

(a) must be located at a pivot.

(b) must be located at the object’s center of gravity.

(c) should be located at the edge of the object.

(d) can be located anywhere.

Two identical, uniform beams are symmetrically set up against each other (Fig. 9–87) on a floor with which they have a coefficient of friction\({\mu _{\rm{s}}} = 0.50\). What is the minimum angle the beams can make with the floor and still not fall?


Question:A 50-story building is being planned. It is to be 180.0 m high with a base 46.0 m by 76.0 m. Its total mass will be about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{\bf{kg}}\)and its weight therefore about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{N}}\). Suppose a 200-km/h wind exerts a force of\({\bf{950}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\)over the 76.0-m-wide face (Fig. 9–80). Calculate the torque about the potential pivot point, the rear edge of the building (where acts in Fig. 9–80), and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building’s face, and that the building is not anchored in bedrock. [Hint:\({\vec F_{\rm{E}}}\)in Fig. 9–80 represents the force that the Earth would exert on the building in the case where the building would just begin to tip.]
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